Find all positive integer solutions of $a^3 + 2b^3 = 4c^3$.
Proof: There don't exist any integer solutions for the give equation. Proof by the Well Ordering Principle.
Let $d$ be the set of all such possible combinations in form $(a, b, c)$.
By the Well Ordering Principle, we that every nonempty set of non-negative integers always has a smallest element. So, there must be a tuple $(a, b, c)$ such that $\max(a, b, c)$ is the least possible value in $d$.
We also know that $a$, $b$ and $c$ must be even. Since $a$ must be even if $4c^3$ has to be even but we know that $4c^3$ is even, so $a$ is the same. Similarly it can be shown that all $a$, $b$ and $c$ are even.
Now, let $a = 2x$, $b = 2y$ and $c = 2z$.
But then, $$a^3 + 2b^3 = 4c^3$$ $$\implies (2x)^3 + 2*(2y)^3 = 4*(2z)^3$$ $$\implies 8x^3 + 16y^3 = 32z^3$$ $$\implies x^3 + 2y^3 = 4z^3$$
So, $(x, y, z) \in d$. But, $\max(x, y, z) < \max(a, b, c)$. But $(a, b, c)$ was supposed to be the smallest such member. We have a contradiction here. There is no smallest $(a, b, c)$ which satisfy this criterion and hence no $(a, b, c)$ at all and $d$ is empty. Hence, there exist no integers $a$, $b$ and $c$ which satisfy $a^3 + 2b^3 = 4c^3$
I wrote this proof. Is it enough to prove the above question? Am I right? Are there any mistakes in the above proof? If there are any, please help me.
Do you have any other ways of proving it?